A **reflexive pair** is a parallel pair $f,g\colon A\rightrightarrows B$ having a common section, i.e. a map $s\colon B\to A$ such that $f \circ s = g \circ s = 1_B$. A **reflexive coequalizer** is a coequalizer of a reflexive pair. A category **has reflexive coequalizers** if it has coequalizers of all reflexive pairs.

Dually, a reflexive coequalizer in the opposite category $C^{op}$ is called a **coreflexive equalizer** in $C$.

Reflexive coequalizers should not be confused with split coequalizers, a distinct concept.

Any congruence is a reflexive pair, so in particular any quotient of a congruence is a reflexive coequalizer.

If $T$ is a monad on a cocomplete category $C$, then the category $C^T$ of Eilenberg Moore algebras is cocomplete if and only if it has reflexive coequalizers. This is the case particularly if $T$ preserves reflexive coequalizers.

This is due to (Linton).

Suppose $C^T$ has reflexive coequalizers. Then $C^T$ certainly has coproducts, because if $A_i$ is a collection of $T$-algebras, then we can form the coequalizer in $C^T$ of the reflexive pair

$\sum_i F U F U A_i
\underoverset
{\sum_i F U \varepsilon A_i}
{\sum_i \varepsilon F U A_i}
{\rightrightarrows}
\sum_i F U A_i$

using the fact that the displayed coproducts exist because, for example,

$\sum_i F U A_i \cong F(\sum_i U A_i)$

since the left adjoint $F$ preserves coproducts, assumed to exist in $C$. That this reflexive coequalizer is the coproduct $\sum_i A_i$ in $C^T$ is routine.

Finally, a category with coproducts and reflexive coequalizers is cocomplete. It suffices that general coequalizers exist, but it is easily seen that if

$f, g \colon A \stackrel{\to}{\to} B$

is a parallel pair, then the coequalizer of the reflexive pair

$A + B \stackrel{\overset{(f, 1_B)}{\to}}{\underset{(g, 1_B)}{\to}} B$

(note both maps are retracts of the inclusion $B \to A + B$) also exists, and gives the coequalizer of the first pair. This completes the proof.

If $F\colon C\times D\to E$ is a functor of two variables which preserves reflexive coequalizers in each variable separately (that is, $F(c,-)$ and $F(-,d)$ preserve reflexive coequalizers for all $c\in C$ and $d\in D$), then $F$ preserves reflexive coequalizers in both variables together.

This is emphatically *not* the case for arbitrary coequalizers.

**of proposition ** Suppose given two reflexive coequalizers

$c_0 \stackrel{\to}{\to} c_1 \to c_2$

$\,$

$d_0 \stackrel{\to}{\to} d_1 \to d_2$

and let $c_{i j}$ denote $F(c_i, d_j)$ so that we have a diagram

$\array{
c_{0 0} & \stackrel{\to}{\to} & c_{0 1} & \to & c_{0 2} \\
\downarrow \downarrow & & \downarrow \downarrow & & \downarrow \downarrow \\
c_{1 0} & \stackrel{\to}{\to} & c_{1 1} & \to & c_{1 2} \\
\downarrow & & \downarrow & & \downarrow \\
c_{2 0} & \stackrel{\to}{\to} & c_{2 1} & \to & c_{2 2}
}$

in which all rows and columns are reflexive coequalizers (using preservation of reflexive coequalizers in separate variables), and all squares are *serially* commutative. According to Toposes, Triples, Theories, lemma 4.2 page 248, the diagonal is also a (reflexive) coequalizer, as claimed. (See also the lemma on page 1 of Johnstone’s Topos Theory.)

Proposition is particularly interesting when $F$ is the tensor product of a cocomplete closed monoidal category $C$. In this case it implies that the free monoid monad on such a category preserves reflexive coequalizers, and thus (by Linton’s theorem) the category of monoid objects in $C$ is cocomplete.

Reflexive coequalizers in Set commute with finite products:

the $n$-fold product functors $Set^n \stackrel{\prod}{\to} Set$ preserve reflexive coequalizers.

This follows from prop. as well as from the fact that the diagram category $\{ 0 \stackrel{\overset{d_0}{\to}}{\stackrel{\overset{s_0}{\leftarrow}}{\underset{d_1}{\to}}} 1\}$ with $d_0 \circ s_0 = d_1 \circ s_0 = id$ is a sifted category.

Of course, the diagonal functor $\Delta: Set \to Set^n$, being left adjoint to the product functor, preserves reflexive coequalizers; therefore the composite

$Set \stackrel{\prod \Delta}{\to} Set: x \mapsto \hom(n, x)$

also preserves reflexive coequalizers.

This has a further consequence which is technically very convenient:

If $T$ is a finitary monad on $Set$, then $T$ preserves reflexive coequalizers.

We have a coend formula for $T$:

$T(-) \cong \int^{n \in Fin} T(n) \times \hom(n, -)$

and since this is a colimit of functors $\hom(n, -)$ which preserve reflexive coequalizers, $T$ must also preserve reflexive coequalizers.

Since finitary monads $T$ preserve reflexive coequalizers, it follows that the monadic functor $U \colon Set^T \to Set$ reflects reflexive coequalizers, and so since $Set$ has reflexive coequalizers, $Set^T$ must as well. Therefore, by proposition , $Set^T$ is cocomplete. This is actually true for infinitary monads $T$ on $Set$ as well, at least if we assume the axiom of choice (see here for a proof), but the argument just given is a choice-free proof for the case of finitary monads.

- Reflexive coequalizers figure in the crude monadicity theorem.

- Fred Linton,
*Coequalizers in categories of algebras*, Seminar on Triples and Categorical Homology Theory, Lecture Notes in Mathematics Vol. 80 (1969), 75-90.

- Michael Barr and Charles Wells,
*Toposes, Theories, and Triples*, Reprints in Theory and Applications of Categories (2005), 1-289. (online pdf)

- Peter Johnstone, Topos Theory, London Mathematical Society Monographs no. 10, Academic Press, 1977.

Last revised on November 30, 2022 at 11:20:46. See the history of this page for a list of all contributions to it.